Difference between revisions of "Problem 4 Old Kiwi" - Rhea

Latest revision as of 15:20, 3 July 2008

SYSTEM 1 - $ h[n] = \delta[n+1] + \delta[n-1] $

A - $ y[n] = \sum_{k= \neg \infty}^\infty x[k]h[n-k] $

$ y[n] = \sum_{k= \neg \infty}^\infty h[k]x[n-k] $

Now, for the output to depend only on the current input value,

$ h[k] = 0 $ if $ k \neq 0 $ $ \rightarrow h[k] = A \delta[k] $ $ \rightarrow h[n] = K \delta[n] $ $ \rightarrow y[n] = Kx[n] $

For an LTI system to be memoryless, the output value of 'n' should only depend on the CURRENT input value of 'n'. That is, $ y[n] = Kx[n] $. But, in this case, the output value of 'n' depends on the past and future values of 'n'. In other words, $ y[n] \neq Kx[n] $. As a result, the system is NOT memoryless or has memory.

B - $ y[n] = \sum_{k= \neg \infty}^\infty x[k]h[n-k] $

To be causal, h[n-k] = 0 for k > n $ \rightarrow $ h[k] = 0 for k < 0

For an LTI system to be causal, the output should NOT depend on the future input values. In other words, $ h[k] = 0 for k < 0 $. But, in this case, the output does depend on the future input values and as a result, is not causal. That is, $ h[k] \neq 0 for k < 0 $. One can easily check the causality of an LTI system by looking at the negative x-axis.

C - Stable (Why?)

x[n] < B for all n

$ y[n] = \sum_{k= \neg \infty}^\infty h[k]x[n-k] $

$ \mid x[k] \mid $ < B for all k

$ \mid y[n] \mid = \mid \sum_{k= \neg \infty}^\infty h[k]x[n-k] \mid $

$ \mid y[n] \mid \leq \mid \sum_{k= \neg \infty}^\infty h[k]x[n-k] \mid $

$ \mid y[n] \mid \leq \sum_{k= \neg \infty}^\infty \mid h[k] \mid \mid x[n-k] \mid $

$ \mid y[n] \mid \leq \sum_{k= \neg \infty}^\infty \mid h[k] \mid \times B $

$ \mid y[n] \mid < \infty $ IFF $ \sum_{k= \neg \infty}^\infty \mid h[k] \mid < \infty $

$ \sum_{k= \neg \infty}^\infty h[k] = \sum_{k = \neg \infty}^\infty {\delta[n+1] + \delta [n-1]} $

$ \sum_{k= \neg \infty}^\infty h[k] = \sum_{k = -1}^1 {1} = 2 < \infty $

Therefore, h[n] is stable.

SYSTEM 2 - $ h(t) = e^t[u(t-2) - u(t-5)] $

A - $ y(t) = \int_{ \neg \infty}^\infty x(t)h(t- \tau) d\tau $

$ y(t) = \int_{ \neg \infty}^\infty h(t)x(t - \tau) d\tau $

Now, for the output to depend only on the current input value,

$ h(t) = 0 $ if $ t \neq 0 $ $ \rightarrow h(t) = A \delta[k] $ $ \rightarrow y(t) = Kx(t) $

For an LTI system to be memoryless, the output value of 't' should only depend on the CURRENT input value of 't'. That is, $ y(t) = Ax(t) $. But, in this case, the output value of 't' depends on the past value of 't'. In other words, $ y(t) \neq Ax(t) $. As a result, the system is NOT memoryless or has memory.

B - $ y(t) = \int_{\neg \infty}^\infty x(t)h(t - \tau) d\tau $

To be causal, h(t) = 0 for t < 0

For an LTI system to be causal, the output should NOT depend on the future input values. In this case, the output only depends on the past input values and as a result, is causal. Therefore, $ h(t) = 0 for t < 0 $. One can easily check the causality of an LTI system by looking at the negative x-axis.

C - Stable (Why?)

x(t) < B for all t

$ y(t) = \int_{\neg \infty}^\infty h(\tau)x(t - \tau) d\tau $

$ \mid y(t) \mid = \mid \int_{\neg \infty}^\infty x(t - \tau)h(\tau) d\tau \mid $

$ \mid y(t) \mid \leq \int_{\neg \infty}^\infty \mid x(t - \tau) \mid \mid h(\tau) \mid d\tau $

$ \mid y(t) \mid \leq B \times \int_{\neg \infty}^\infty \mid h(\tau) \mid d\tau $

$ \mid y(t) \mid < \infty $ IFF $ \int_{\neg \infty}^\infty \mid h(\tau) \mid d\tau < \infty $

$ \int_{\neg \infty}^\infty h(\tau) d\tau \leq \int_{\neg \infty}^\infty e^\tau[u(\tau-2) - u(\tau-5)] d\tau $

$ \int_{\neg \infty}^\infty h(\tau) d\tau \leq \int_{2}^5 {e^\tau} d\tau $

$ \int_{\neg \infty}^\infty h(\tau) d\tau \leq {e^5 - e^2} = (Constant) < \infty $

Therefore, h(t) is stable.

Alternaive Solutions

Problem 4 - Mistake in solution posted_Old Kiwi